Question

The electric potential in some region is given by V=6x-8xy^2-8y+6yz-4z^2.Find force on 2C charge placed at origin

Answer 1

First of all we have to find out electric field .
Electic field is given by $\bold{E=-\frac{\delta{V_x}}{\delta{x}}i-\frac{\delta{V_y}}{\delta{y}}j-\frac{\delta{V_z}}{\delta{z}}k}$

So, Ex = -δVx/dx = -δ(6x -8xy² -8y +6yz - 4z²)/δx
= -6 - 8y² - 0 + 0 - 0 , put (0,0,0)
Ex = -6

Similarly
Ey = -δVy/δy = -(-16x - 8 + 6z - 0) , put (0,0,0)
Ey = 8

Ez = -δVz/δz = (0 - 0 - 0 + 6y - 8z), put (0,0,0)
Ez = 0

Hence, E = -6i + 8j + 0k
|E| = 10 N/C

We know, Force = charge × electic field
e.g., F = qE
so, F = 2 × 10 = 20N

Hence, answer is 20N

Answer 2

Explanation:

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